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Question 744406: write an equation for the conic section
parabola with vertex at(0,0) and directrix x = 5
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! write an equation for the conic section
parabola with vertex at(0,0) and directrix x = 5
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Form: (x-h)^2 = 4p(y-k)
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The distance from the vertex to the directrix is p = -5
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Substitute for h,k and p to get:
(x-0)^2 = -20(y-0)
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x^2 = -20y
y = (-1/20)x^2
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Cheers,
Stan H.
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Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! write an equation for the conic section
parabola with vertex at(0,0) and directrix x = 5
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This is a parabola that opens leftward.
Its basic form of equation: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
given vertex: (0,0)
axis of symmetry: y=0 or x-axis
p=5 (distance from vertex to directrix on the axis of symmetry)
4p=20
Equation: y^2=-20x
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